# Verifying Woronowicz’s proof that all ${\text{SU}_{q}}(2)$’s are isomorphic as $C^{*}$-algebras, where $-1 < q < 1$

This question is related to one that I asked some time ago.

Definition 1. Let $q \in (-1,1)$. Let $A$ be a unital $C^{*}$-algebra and $(x,y)$ a pair of elements of $A$. Then define the $q$-relations for $(x,y)$ to be the following five relations: \begin{align} x^{*} x + y^{*} y & = 1_{A}. \\ x x^{*} + q^{2} y y^{*} & = 1_{A}. \\ y^{*} y & = y y^{*}. \\ x y & = q y x. \\ x y^{*} & = q y^{*} x. \end{align}

Definition 2. The compact quantum group ${\text{SU}_{q}}(2)$ is defined as the universal unital $C^{*}$-algebra that is generated by a pair $(x,y)$ of elements satisfying the $q$-relations. We shall call $(x,y)$ ‘a generating pair for ${\text{SU}_{q}}(2)$’.

Woronowicz’s proof that all the ${\text{SU}_{q}}(2)$’s are isomorphic as $C^{*}$-algebras (contained in his paper Twisted $\text{SU}(2)$ Group. An Example of a Non-Commutative Differential Calculus) goes as follows:

1. Let $q \in (-1,1)$.

2. Let $(a,b)$ and $(\alpha,\beta)$ be generating pairs for ${\text{SU}_{q}}(2)$ and ${\text{SU}_{0}}(2)$ respectively.

3. The pair (of elements of ${\text{SU}_{q}}(2)$) $$\phi_{(a,b)} \stackrel{\text{df}}{=} (a ~ f(a^{*} a),b ~ g(b^{*} b)),$$ obtained via the continuous functional calculus, satisfies the $0$-relations, where $f,g \in C([0,1])$ and have the following properties: $$f(t) = \begin{cases} 0 & \text{if  t = 0 }, \\ \dfrac{1}{\sqrt{t}} & \text{if  t \in \left[ 1 - q^{2},1 \right] }; \end{cases} \qquad g(t) = \begin{cases} 0 & \text{if  t \in \left[ 0,q^{2} \right] }, \\ 1 & \text{if  t = 1 }. \end{cases}$$ Then by the universal property of ${\text{SU}_{0}}(2)$, there exists a unique morphism $\Phi: {\text{SU}_{0}}(2) \to {\text{SU}_{q}}(2)$ such that $(\Phi(\alpha),\Phi(\beta)) = \phi_{(a,b)}$.

4. The pair (of elements of ${\text{SU}_{0}}(2)$) $$\psi_{(\alpha,\beta)} \stackrel{\text{df}}{=} \left( \sum_{n = 1}^{\infty} \sqrt{1 - q^{2 n}} \left[ (\alpha^{*})^{n - 1} \alpha^{n} - (\alpha^{*})^{n} \alpha^{n + 1} \right], \sum_{n = 0}^{\infty} q^{n} (\alpha^{*})^{n} \beta \alpha^{n} \right)$$ satisfies the $q$-relations. Then by the universal property of ${\text{SU}_{q}}(2)$, there exists a unique morphism $\Psi: {\text{SU}_{q}}(2) \to {\text{SU}_{0}}(2)$ such that $(\Psi(a),\Psi(b)) = \psi_{(\alpha,\beta)}$.

5. As $\Phi \circ \Psi = \text{Id}_{{\text{SU}_{q}}(2)}$ and $\Psi \circ \Phi = \text{Id}_{{\text{SU}_{0}}(2)}$, one can conclude that ${\text{SU}_{q}}(2) \cong {\text{SU}_{0}}(2)$.

Question 1. How can we derive the infinite series in Step 4 and prove that the pair $\psi_{(\alpha,\beta)}$ satisfies the $q$-relations? I believe that once this is answered, Step 5 can be tackled.
I must also mention that Woronowicz claims the following identity: $$\sum_{n = 1}^{\infty} \sqrt{1 - q^{2 n}} \left[ (\alpha^{*})^{n - 1} \alpha^{n} - (\alpha^{*})^{n} \alpha^{n + 1} \right] = \sum_{n = 0}^{\infty} \frac{\left( 1 - q^{2} \right) q^{2 n}}{\sqrt{1 - q^{2 n}} + \sqrt{1 - q^{2 (n + 1)}}} (\alpha^{*})^{n} \alpha^{n + 1}.$$ The infinite series on the right is readily seen to be absolutely convergent. I could easily verify that \begin{align} \forall N \in \Bbb{N}: \qquad & ~ \sum_{n = 1}^{N} \sqrt{1 - q^{2 n}} \left[ (\alpha^{*})^{n - 1} \alpha^{n} - (\alpha^{*})^{n} \alpha^{n + 1} \right] \\ = & ~ \sum_{n = 0}^{N - 1} \frac{\left( 1 - q^{2} \right) q^{2 n}}{\sqrt{1 - q^{2 n}} + \sqrt{1 - q^{2 (n + 1)}}} (\alpha^{*})^{n} \alpha^{n + 1} - \sqrt{1 - q^{2 N}} (\alpha^{*})^{N} \alpha^{N + 1}, \end{align} but I was unable to prove that $\displaystyle \lim_{N \to \infty} \sqrt{1 - q^{2 N}} (\alpha^{*})^{N} \alpha^{N + 1} = 0_{{\text{SU}_{0}}(2)}$.
Question 2. How can we prove that $\displaystyle \lim_{N \to \infty} \sqrt{1 - q^{2 N}} (\alpha^{*})^{N} \alpha^{N + 1} = 0_{{\text{SU}_{0}}(2)}$?